of 1 62
Six-fold Reality
By
Ian Beardsley
Copyright © 2023 by Ian Beardsley
of 2 62
Introduction……………………………………………………………………..3
The Theory……………………………………………………………………….4
Conclusion……………………………………………………………………….29
Continuing On………………………………………………………………….32
The Cosmic Calendar…………………………………………………………43
Appendix 1………………………………………………………………………..46
Appendix 2……………………………………………………………………….50
Appendix 3……………………………………………………………………….55
Appendix 4……………………………………………………………………….60
of 3 62
Introduction
Here I show that the physical characteristics of the Universe can be described by the six-fold
unfolding of space and time in relatively simple terms, namely the characteristics inertia, proton
radius, proton charge, and the Sun’s magnetic field. Further that to do this we need to formulate
a constant that bridges the macrocosmos to the microcosmos, that is the planets to the proton
and the elements. Interestingly, their seems to be correspondence of this with the Old
Testament, which is discussed in terms of the cause being perhaps in the Universality of the six-
fold. The six-fold theory of reality results in a cosmic calendar, and it is correlated the Earth’s
past extinction events. It predict we could be in one now plus or minus 20 million years.
However many scientists believe we are in one now (Called the Anthropocene) where human
impact on the Earth is changing it drastically, mostly due to burning fossil fuels and putting CO2
into the atmosphere in excess of 350 parts per million, which causes global warming. We can
change that future, however, by switching to cleaner, renewable energy sources. We just have to
choose it.
of 4 62
The Theory Years ago I saw a t-shirt that said:
God said Let there be light
And there was light.
It is one of Maxwell’s Equations, the one explaining that light (where c is the speed of light) is an
electromagnetic wave.
I was working on a theory for matter, space, and time, where matter, or inertia, was a property
of space and time, as well as the gravitational and electric fields, and that this was based in six-
fold symmetry. I sought not just to describe the microcosmos, the small atoms and fundamental
particles from which matter is made, but to describe the macrocosmos, the planets, Sun, and
moon, in terms of one another through some constant k that connected these two realms which
would be represented on the microscale by Planck’s constant h, and on the macroscale by the
universal constant of gravitation G. I had success with this and the work can be found in my
paper and book The Six-fold Nature of Reality, though we go into much of it here.
While working on this I found something very interesting. I was considering the Moon and its
kinetic energy and the Earth and its kinetic and six rotations of the Earth for my
theory of six-fold symmetry, and found that
Equation 1.
Equation 2.
Where
Equation 3.
Is a constant of proportionality and is the surface area of a proton, and is the rest energy
of a proton. is the fine structure constant. I found this interesting because aside from the
Bible’s Let there be light, and there was light, the light speed of light c squared is divided into 6
days, and we all know the Bible says God created the heaven’s and the Earth in six days, and
rested on the seventh. That is divided into 6 days it is the kinetic energy of the Moon to the
kinetic energy of the Earth by a factor of the fundamental particle from which matter is built, the
proton, its surface area and rest energy. The equation is approximate but it is closer to six days
than it is to five days or seven days. It is most accurate if you compute the kinetic energies of the
× 𝔹 =
1
c
(
4π𝕁 +
∂𝔼
t
)
K E
m
K E
e
1
α
2
h
Gc
4π r
2
p
E
p
K E
e
K E
m
6d a ys
c
2
κ
Pr otonSur fac Area
Pr oton RestEnerg y
Ear thKin et icE nerg y
Moon KineticEnerg y
=
6d a ys
LightSpeedSqu ared
4π r
2
p
E
p
α
of 5 62
Moon and the Earth using lunar orbital velocity at aphelion and the Earth orbital velocity at
perihelion (93% accurate). The six days comes from rounding to the nearest integer (6) because
the theory in which I was developing the idea was based on nearest integers as is done in
chemistry where you cannot have fractional neutrons when determining the number of neutrons
an element has from the molar mass and atomic number (See Appendix 1).
The equation comes from my equation for proton-seconds that relates the components of the
molecular skeletons of life hydrogen and carbon that make up the hydrocarbons, to time, and
which ultimately predicts the radius of a proton. The equations are:
Equation 4.
That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:
Equation 5.
For time t greater than 6 seconds we have fractional protons. For t<1 we have the elements
heavier than carbon. 6 seconds is hydrogen, one second is carbon. I stumbled upon our
equations 1 and 2 which are what this paper is highlighting, when trying to eliminate time, t, in
equations 4 and 5. Indeed we see 6 seconds so closely gives 1 proton (hydrogen) and 1 second so
closely gives 6 protons (carbon) that we are compelled to look at the possibility that the duration
of a second is a natural unit that we accidentally created. The unit of a second came about from
dividing the day into 24 hours and that into 60 minutes and that into 60 seconds. We divided
the hour into 60 minutes and the minute into 60 seconds because our ancestors, the Sumerians,
and Babylonians, and from them the Ancient Greeks, chose to divide time by 60 because the 60
is so divisible evenly (evenly divisible by 1, 2, 3, 4, 5, 6, 8, 10, 12, 15 30,…). So that is something
natural that could be making the unit one second so functional in equations 4 and 5. Since there
are further about 30 days in a month, and 365.25 days close to the 360 degrees in a circle giving
the earth moving through about one degree a day, I made a first guess at the second as natural
by considering the the second as given by the Earth day by a factor of the kinetic energy of the
Moon and the the kinetic energy of Earth, which panned out nicely, I wrote:
Equation 6.
It is actually equal to 1.2 seconds. But this was not good enough because I sought a theory for
inertia the property that defines mass, from those constants which measure the properties of
space and time at both the microlevel (which would be Planck’s constant h) and and at the
macrolevel (which would be the universal constant of gravitation G). I stumbled upon a very
good lead in the work of Warren Giordano, which panned out nicely.
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton 6secon d s = hydr ogen(H )
K E
moon
K E
earth
(Ear th Da y) 1secon d
of 6 62
He wrote in his paper The Fine Structure Constant And The Gravitational Constant: Keys To
The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where
’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
I found I could eliminated the and at the same time get the six of the six-fold symmetry
with which I was working by considering Avogadro’s number .
I suggested there exists some k that serves as a constant that describes both the microcosmos
and macrocosmos from the proton, to the atoms, to planetary orbits. It is such that the square
root of it times the earth orbital velocity is 6, because we are guessing we are dealing with six-
fold symmetry as the basis of Nature. That is
Eq. 7
We have that k is
Eq. 8
This follows from what Warren Giordano noticed that
Eq. 9
Without the right units. I noticed since Avogadro’s number is that I
could introduce an equation of state for the periodic tables of the elements:
Eq. 10
Eq. 11
Let us say we were to consider Any Element say carbon . Then in general
We have
and
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
h
1 + α
α
10
23
10
23
6.02E 23atom s
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
h(1 + α) 10
23
= G
6.02 × 10
23
6 × 10
23
= 1
gra m
atom
h
(1 + α)
G
N
A
H = 6.0003
kg
2
s
m
𝔼
h
(1 + α)
G
N
A
𝔼 = 6.0003kg
2
s
m
=
6gra m s
6protons
N
A
=
6(6E 23pr oton s)
6gra m s
of 7 62
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And,
Therefore we always have:
is a variable, the number of protons in multiplied by Avogadro’s number.
Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Equation. 12
While we have masses characteristic of the microcosmos like protons, and masses characteristic
of the macrocosmos, like the upper limit for a star to become a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 — More mass than that and she will collapse — we do not
have a characteristic mass of the intermediary world where we exist, a truck weighs several tons
and tennis ball maybe around a hundred grams. To find that mass let us take the geometric
mean between the mass of a proton and the mass of 1.44 solar masses. We could take the
average, or the harmonic mean, but the geometric mean is the squaring of the proportions, it is
the side of a square with the area equal to the area of the rectangle with these proportions as its
sides. We have:
Equation. 13
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Equation 14
All we really need to do now is divide equation 12 by equation 14 and we get an even number
that is the six of our six-fold symmetry.
N
A
= 6E 23
h
(1 + α)
G
N
A
= 6.0003kg
2
s
m
𝔼
N
A
=
Z 6E 23protons
Z gr a m s
N
A
=
Z 6E 23protons
Z gr a m s
N
A
𝔼 = 6E 23
N
A
𝔼
h
(1 + α)
G
N
A
𝔼 v
e
= 422.787kg
M
= 1.98847E 30kg
m
p
= 1.67262E 27kg
m
i
= (2.8634E 30)(1.67262E 27) = 69.205kg
of 8 62
Equation. 15
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Eq. 16
Where k is a constant, given
Eq. 17
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the electron degeneracy pressure and collapse. The non-
relativistic equation is:
Let us approximate 0.77 with 3/4. Since we have our constant
Equation. 18
Equation 19
Then
Equation 20
Since our constant k in terms the Chandrasekhar limit is
Equation 21
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared, that is it represents the ground state. It is
1
m
i
h
(1 + α)
G
N
A
𝔼 v
e
= 6.1092 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
773.5
s
m
M 0.77
c
3
3
G
3
N
m
4
p
= 1.41
k =
1
m
2
i
h
1 + α
G
N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
3
G
3
m
2
p
1/2
= h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
of 9 62
Since
We are suggesting the earth orbit is the ground state for our planetary system. We suggest it
holds for any planetary system because k as we will see is a natural constant that solves many
physical problems on many levels, not just planetary systems but atomic systems and the
particles that make them up.
Let us now recall equations 4 and 5
While we have considered them to be proton-seconds because they are a mass divided by the
mass of a proton, we can consider these two masses to cancel and say they are equal to 1 second
and six seconds respectively. We have that carbon, which is to evaluate them at one second, is
the radius of a proton:
Equation 22.
This gives the radius of a proton is:
Equation 23.
Where . The experimental value of the proton radius is 0.833fm+/-0.014fm
Radius of hydrogen atom
Equation 24
α
2
=
U
e
m
e
c
2
k v
e
= 6
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton 6secon ds = hydr ogen(H )
1
α
2
m
p
h 4π r
2
p
Gc
= 6secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
= 1secon d
R
H
= 1.2E 10m
t =
R
h
v
= R
h
1
α
2
m
p
h 4π
Gc
of 10 62
Remember our constant k equation 21 (I find we have to divide by two somewhere and I think
this is because we are looking at packing protons here so they are offset by half their radius from
one another):
Equation 25
And 1 second in terms of the atom is given by
Then the equation for the radius of a proton is from equation 24:
Equation 26
Let’s verify our equation:
Making the approximation 9/8~1 we can write equation 26 as (We suggest we have picked up
the fraction 9/8 by making several approximations):
Equation 27
Which gives
We form constants:
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
1 + α
1
N
A
𝔼
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
N
A
𝔼
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12secon d s
3 2
16
h
Gπ
1
α
2
m
p
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
9
8
2
1
m
p
hc
4π
3
G
R
H
N
A
𝔼
r
p
=
9
8
2
1
1.67E 27
(6.626E 34)(299,792, 459)
(6.674E 11)4π
3
1.2E 10
6.02E 23
= 0.93f m
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
= 8.26935E 16m 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
=
1
kg
(
kg
)
m
of 11 62
Equation 28
Equation 29
And we have the Equation:
Equation 30
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
Equation 31
We have successfully predicted the radius of a proton with a value for t in terms of the natural
constants. Let us return to equation 1:
Which we wrote as equation 2:
Using rather the second as
Which suggests the Moon plays some mysterious role in Nature. Indeed there is something
mysterious surrounding the Moon. If in the planet that has life orbiting a star there is an
indication to its intelligence that there is a mystery before it, it is that for our star the Sun, the
moon perfectly eclipses it as seen from the Earth. This is because:
Which are approximately equal. As well we can look at it as:
k =
hc
2π
3
G
= 6.93E 9kg
R
H
N
A
𝔼
= 1.99E 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
R
H
N
A
𝔼
1
α
2
h
Gc
4π r
2
p
E
p
K E
e
K E
m
6d a ys
c
2
κ
Pr otonSur fac Area
Pr oton RestEnerg y
Ear thKin et icE nerg y
Moon KineticEnerg y
=
6d a ys
LightSpeedSqu ared
K E
moon
K E
earth
(Ear th Da y) 1secon d
(lun ar orbit)
(ear th orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lun ar ra diu s)
solar r a dius
=
1,738.1
696,00
= 0.0025
of 12 62
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver (Ag)
by molar mass, these elements being used for religious and ceremonial jewelry since ancient
times, is the same 1.8:
The sun is gold in color, the moon is silver in color. Thus the mystery is in that the moon is 400
times smaller than the Sun, but 400 times further from the Sun than it is from Earth and the
Solar Radius is 1.8 times the lunar orbital radius.
Indeed equation 22, which was
Tells us one second is in exactness for a value of time approximating a second;
Equation 32.
Which is very close to exactly one second, whereas
Equation 33.
It is at this point that we list the values of our constants that we are using and that we will be
using:
(lun ar ra diu s)
(lun ar orbit)
=
(1,738.1)
(384,400)
= 0.00452
solar r a dius
ear th orbit
=
696,000
149,597,870
= 0.00465
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
1
α
2
m
p
h 4π r
2
p
Gc
= 6secon d s
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
K E
m
K E
e
(Ear th Da y) = 1.2secon ds
of 13 62
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
And we want to look at how we formulated proton-seconds and how they are explaining the
nature of matter in terms of space and time.
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 34
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 35
Which describes mass per meter over time, which is:
Equation 36
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
Equation 37
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 38
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 39
m
P
: 1.67262 × 10
27
kg
h : 6.62607 × 10
34
J s
r
p
: 0.833 × 10
15
m
G: 6.67408 × 10
11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
G = 6.674E 11N
m
2
s
2
= 6.674E 11
m
3
s
2
kg
h
Gc
=
kg s
m
1.82E 16
kg s
m
α
1
α
2
(1.82E 16kg s /m)(18,769) = 3.416E 12kg s /m
α
2
=
U
e
m
e
c
2
S
p
S
p
= 4π r
2
p
= 8.72E 30m
2
of 14 62
We take the square root to get meters:
Equation 40
We multiply that with the value we have in equation 37:
Equation 41
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 42
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 43
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 44
Equation 45
If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of
the elements in the periodic table occur for dividing by something less that 1-second.
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. Since Planck’s constant h is a
measure of energy over time where space and time are concerned it must play a role. Of course
the radius of a proton plays a role since squared and multiplied by it is the surface area of our
proton embedded in space. The gravitational constant is force produced per kilogram over a
distance, thus it is a measure of how the surrounding space has an effect on the proton giving it
inertia. The speed of light c has to play a role because it is the velocity at which events are
separated through time. The mass of a proton has to play a role because it is a measurement of
inertia itself. And alas the fine structure constant describes the degree to which these factors
have an effect. We see the inertia then in equation 6 is six protons over 1 second, by dimensional
analysis.
S
p
= 2.953E 15m
(2.953E 15m)(3.416E 12
kg s
m
) = 1.009E 26kg s
m
p
m
h
=
1E 26
6secon d s
= 1.67E 27kg
m
c
=
1E 26kg s
1secon d
= 1E 26kg = 6proton s = 6m
p
m
p
=
1E 26kg s
6secon d s
m
c
=
1E 26kg s
1secon d
4π
of 15 62
We can make a program that finds integer solutions of
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
1
α
2
m
p
h 4π r
2
p
Gc
Fig. 1
of 16 62
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
Why did I write equation 23
Using 18/3 instead of 6? It is to emphasize the dynamics of six-fold symmetry.
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
of 17 62
The periodic table is periodic over 18 groups, which means if we proceed by number of protons
in an element, and start over after every 18 elements, the elements group themselves by their
natural properties, which is at the basis of doing chemistry and requires an entire textbook on
chemistry to go into fully, but here we need only point out that
The periodic table is periodic over 18 groups. 2, 3 are the smallest primes, the lowest factors in
which we can factor numbers, we have for the dynamics of six-fold symmetry:
In light of this we ask why six days (six rotations) of the Earth in equation 1:
The bible begins with God said let there by light and there was light, and it there goes into God
created the heavens and the earth in six days and rested on the seventh. Did someone in that
Middle Eastern desert eat a hallucinogenic plant and have a vision wherein he saw the
connection of the moon and earth with light and the six days of creation, or did he choose 6 days
for creation because he knew six was dynamic. Indeed the Bible speaks of the ratio of the
circumference of a circle to its diameter being 3. It is more accurately 3.141 but three is a close
estimate and it is not trivial to go on calculating the digits after the decimal. In fact the Ancient
Greek method of calculating pi begins with the estimate that the perimeter of a regular hexagon
to its diameter is three. The regular hexagon has six-sides and the property that its radius is
equal to its sides, and this is the power of six-fold symmetry because if its radius is one, each
side is one, of which there are six, and twice the radius is its diameter. Thus with six sides of one
each, the perimeter is 6, and the diameter is two giving pi=6/2=3. This regular hexagon
inscribed in a circle approximates a circle in all of its components, so thus approximates pi. If
the Bible says that the world was created in six days, perhaps that was because the regular
hexagon has six sides each equal in length. The seven day week came from the Romans 2000
years ago, the Old Testament was written 3500 years ago in Babylonian captivity. Since the
Moon goes around the earth in about 27 days, then if you are going to divide that into four units,
the week, the week is going to be 7 days. Seven times 4 is the approximately 29 day month, and
2 3 = 6
3 3 = 9
2 9 = 18
3 6 = 18
κ
Pr otonSur fac Area
Pr oton RestEnerg y
Ear thKin et icE nerg y
Moon KineticEnerg y
=
6d a ys
LightSpeedSqu ared
of 18 62
since the Moon goes around the Earth about 12 times in the time the earth goes around the Sun
once, there are 12 months in a year.
Indeed we can say the orbital velocities of the planets are all six if we give them in their natural
units. The natural units are if their orbits are approximately circular, and we call the radii of
such circular orbits as one in each case, we have a circumference of . Since is
approximately 3, then . The orbital velocities in each case are
where .
Thus we want to know the universal gravitational constant in these units:
We can immediately put this to work. In order for the earth to stay in orbit its centripetal force
must equal the gravitational force. Its orbital velocity must be given by:
If G=40, M, the mass of the sun equal to 1, and the orbital radius R is 1, then
Recall equation 7:
Then in terms of the constant k, all of the planetary orbits are given by the orbital velocity of
Earth, . At this time we return to equations 32 and 33:
2π r
π
2π r = 2(3)(1) = 6
2π r /T = 6/1 = 6
T = Pla n et sYear = 1
G = 6.67408E 11
m
3
kg s
2
G = 6.67408E 11
m
3
kg s
2
AU
3
3.3479E 33m
3
1.98847E 30kg
M
9.9477E14s
2
year
2
G = 39.433
AU
3
M
year
2
G 40
AU
3
M
year
2
GM
= 40
AU
3
year
2
G
Mm
R
2
=
mv
2
r
v =
GM
R
v =
(40)(1)
1
= 6
k v
e
= 6
v
e
of 19 62
For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to the
moon. The days become longer by 0.0067 hours per million years. Our Equation
Is actually 1.2 seconds.
We have
24-20=0.0067t
t=597 million years
This was when the earth went through a dramatic change and there was a big explosion of life
(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went
extinct 65 million years ago giving small mammals a chance to evolve paving the way for
humans.
24-x=0.0067t
x=23.5645 hours
We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic
calendar:
What is the next term?
20+3+0.57735+0.4714=24hours
Which bring us to today.
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
K E
m
K E
e
(Ear th Da y) = 1.2secon ds
K E
moon
K E
earth
(Ear th Da y) 1secon d
24h ours
1.2
= 20h ours
3cos(0
) +
2
3
cos(30
) = dinosaur ex t in ct ion =
20hrs + 3hrs +
3
3
hrs +
2
3
=
of 20 62
We have discussed inertia and the proton, but we still need to discuss electric charge and the
solar magnetic field.
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 46
x
3
(x, y, z)
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
ct
2
)
i
··
x (ct
1
ct
2
)
i
Fig. 2
of 21 62
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
Equation 47
We get
x
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
ct
2
)
i
g m =
m
s
2
g =
1
s
2
=
1
t
2
1
s
2
m
s
s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E 34)(4π)(0.833E 15)
2
(6.674E 11)(299,792, 459)
= 5.37E 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E 31) = 9.1435E 19C
of 22 62
Equation 48
We model the formation of the solar system from a slowly rotating gas cloud, a nebula of
gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on
and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,
like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from
which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three
factors that determine its structure, the inward gravity, the pressure gradient outward which
balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The
flattened rotating disc is broken up into rings each that has a mass spread out over it from which
the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth
masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass
spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars
masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the
Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of
about 350 because it is mostly iron condensations with incomplete silicon condensations.
Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is
-1.5 so that the density distribution of the protoplanetary disc is:
Giving a mass
With pressure gradients playing the key role in the formation of solar system, less attention is
payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of
this aspect found something very interesting concerning it. He was Alfven (1942). At the time
people were suggesting instead of the solar system forming from a rotating nebula, rather the
sun came into existence not at the same time at the center of the disc, but rather passed through
clouds and captured material after already existing. He figured for the captured material its
inward component v, and density , at a distance r from the sun, had to conserve mass, which
required:
He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions
would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the
magnetic field becomes important. He considered for simplicity the solar magnetic field was
generated by a dipole moment , a vector quantity, and that a particle moving in the plane of
that vector with mass m and charge q, would have all of both the gravitational and magnetic
forces in that plane, so the problem becomes two-dimensional and required only the and , of
polar coordinates. The differential equations of its motion would be:
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71pr oton s 6pr oton s
σ (r) = σ
0
r
3/2
σ
0
= 3300
M =
2π
0
r
h
r
s
σ (r)rdrdθ
ρ
dM
dt
= 4π r
2
ρv
μ
θ
r
of 23 62
Equation 49
And,
Equation 50
We can integrate equation 50 with the boundary condition that the angular momentum of the
particle is zero at large distances from the sun to get (Appendix 2):
Equation 51
And substitute it into equation equation 49 for to get
Equation 52
Which we can write
Equation 53
We then integrate this with respect to r with the boundary condition that at large r and get
Equation 54
He then notices there is another value for which . It is
Equation 55
This is interesting because it means the particle can never approach the Sun closer than this
value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this
as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the
magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact
that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the
region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen
and helium.
Certainly today we don’t see the planets as having formed from material gathered by the Sun in
its journey, but rather think the Sun and planets formed at the same time from a cloud that
collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through
clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing
clouds from falling into their star. But we can also apply his equation to our Sun today, for which
we know a great deal about its magnetic field, which also happens to be an important thing to
study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels
m
··
r =
GM
m
r
2
+
qμ
·
θ
r
2
+ mr
·
θ
2
m
r
d(r
2
·
θ )
dt
=
qμ
·
r
r
3
mr
2
·
θ =
qμdr
r
2
=
qμ
r
·
θ
m
··
r =
GM
r
2
+
2q
2
μ
2
m
2
r
5
·
r
d
·
r
dr
=
GM
r
2
+
2q
2
μ
2
m
2
r
5
·
r = 0
·
r
2
=
2GM
r
q
2
μ
2
m
2
r
4
·
r = 0
r
c
=
(
q
2
μ
2
2GM
m
2
)
1/3
r
c
of 24 62
with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic
field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict
solar maximums, so we have warning as to whether there will be a magnetic storm that will
knock out our electrical grid and internet, ahead of time.
During solar minimum the solar magnetic field has closed lines, that flow out one pole and into
the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so
that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The
total magnetic field of the Sun at the Earth, is all the components taken together, which are
, , and . The important component is , because it runs north-south, so it is
perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the
component that interacts with the Earth magnetosphere, and when it points southward, it will
connect with the Earth’s magnetosphere which points northward so the solar poles flow into the
Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain
down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t
always stay around the Sun itself, but the solar wind carries it through the solar system until it
collides with the interstellar medium reaching the heliopause. Thus the Sun creates the
Interplanetary Magnetic File (IMF) which has a spiral shape because the Sun rotates once about
every 25 days. But the upshot is that at Earth we have
Moderate Magnetic Field: 10 nT
Strong Magnetic Field: 20 nT
Very Strong Magnetic Field: 30 nT
For our purposes we want to return to equation 55:
B
t
B
x
B
y
B
z
B
z
of 25 62
And ask just what is , because in the time that Alfven was working we worked with magnetic
fields differently, aside from his equation uses a trick, which we still use today, and that is to
consider the magnetic field a dipole. To consider it like this is to say there are two monopoles
opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,
though they are predicted by some modern theories, they have never been found. The trick is in
that by treating the North magnetic pole and South magnetic pole as separate magnetic charges
is to treat them like we do electric charges, the charge of a proton and the charge of an electron,
which can be convenient for making computations, but don’t exist that we know of. So we will
solve equation 56 for , and see what its units are so we can understand what it represents and
we will let m be the mass of a proton and q the charge of a proton. We get:
Equation 56
We can write these units as, by taking Coulombs (C) equal to
Equation 57
This is units of force per current density, which makes sense because a flowing current creates a
force. We can also write it:
Equation 58.
Which is energy per magnetic field strength in that the SI units of magnetic field strength is
amps per meter. This tells us:
Equation 59
Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and
make it a proton:
We have
, , ,
r
c
=
(
q
2
μ
2
2GM
m
2
)
1/3
μ
μ
r
3
GM
m
2
p
q
2
p
= μ =
m
3
kg
C s
a mp secon d s
kg
m
s
2
m
2
a mps
kg
m
2
s
2
m
a mps
μ =
Energ y
Magnet icFiel d St rength
H H
+
+ e
= 1pr oton = 2.18E 18J
q
p
= 1.6E 19C
m
p
= 1.67E 27kg
G = 6.67408E 11N
m
2
kg
2
M
= 1.989E 30kg
of 26 62
We want to look at the Sun as having a current flowing around its equator in a loop with its
radius
We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger
than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar
magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979m):
Equation 60
Equation 61
Or,…
Equation 62
The radius of a proton is 0.833E-15m. We have that r is:
Equation 63
R
= 6.957E 8m
I
1AU
=
5.5362E12amps
1.496E11m
= 37.0a m ps /m
μ =
Ioni zat ionE nerg y
Magnet icFiel d St rength
=
2.18E 18J
37.0A mps /m
= 5.892E 20
J m
A
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
r
r
p
= 5.92 6Pr oton Ra dii = carbon
of 27 62
Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it
provided for by the Sun’s magnetic field.
of 28 62
Let us write the computation as one equation, and verify it. We have
Where
=Ionization energy of Hydrogen
=3.47E-39 (correct)
Equation 64
(0.000034574)(3.47E-39)=1.19972E-43
1.19972E-43^(1/3)
=4.932E-15
So as you can see equation 62 is correct. It says that carbon, the basis of life is in the ratio of the
solar magnetic field and the solar gravitational field.
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
I
=
2B
R
μ
0
I
1AU
=
5.5362E12amps
1.496E11m
= 37.0a m ps /m
1AU = r
e
μ =
Ioni zat ionE nerg y
Magnet icFiel d St rength
=
2.18E 18J
37.0A mps /m
= 5.892E 20
J m
A
IE
H
I
1AU
=
2B
R
μ
0
1
r
e
μ
2
= (5.982E 20)
2
= 3.47E 39
μ
2
=
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
=
(2.18E 18)
2
(12.56637E 7)
2
(1.496E11)
2
4(5E 3)
2
(6.957E8)
2
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
6r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.496E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
of 29 62
Conclusion
Indeed we have shown that we can predict the radius of a proton and its mass in terms of the
properties of space:
Which gives
And the interesting thing is they do not yet have an equation for the radius of a proton, yet we
have it here and it doesn’t just use Planck’s constant, h, of the indeterministic microcosmos, but
uses the gravitational constant, G, of the deterministic macrocosmos. We have further shown
our theory of inertia is based on six-fold symmetry is centered around hydrocarbons, the
skeletons of life chemistry
We find 1 second gives
We find six seconds gives 1 proton is hydrogen:
We further find this same sixfold symmetry describes not just proton systems (atoms) but our
solar system
If is the orbital radius of the moon, and is its orbital period, , Venus and so on. We
also have that the Earth orbit is the ground state for the solar system:
Where,
And
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
= 8.26935E 16m 0.827f m
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton 6secon ds = hydr ogen(H )
(
6 6 6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
r
m
T
m
r
v
T
v
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
of 30 62
We further find that for the moon
Where six days is six rotations of the Earth. This shows the earth/moon/sun system and the
proton are structured around one another in terms of six-fold symmetry. Since there are 360
degrees in a circle, and the earth year is close to 360 rotations of the earth (365.25) the earth
orbit can be divided into six sixty degree sectors, the earth moving through about one degree a
day in its orbit.
We have further found that the solar magnetic field follows through with the same six-fold
symmetry, accommodating carbon:
We also have that
Which is to say
Where
In our equation
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71pr oton s 6pr oton s
1
α
2
m
p
h 4π r
2
p
Gc
M
e
M
m
v
2
e
v
2
m
= 6d ays
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
1
α
2
h
Gc
4π r
2
p
E
p
K E
e
K E
m
6d a ys
c
2
κ
Pr oton Sur fac Area
Pr oton RestEnerg y
Ear thKinet icEnerg y
Moon KineticEnerg y
=
6d a ys
LightSpeedSqu ared
κ =
1
α
2
h
Gc
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton secon d s = carbon(C )
of 31 62
In terms of the Moon the radius of the proton is:
Where . The experimental value of the proton radius is 0.833fm+/-0.014fm. We
can also for a constant k, different than our former constant for which we can write
And we have the Equation:
In our equation
Anything less than a second is heavier than carbon, the core element of life, and the unit of a
second in this is given by the moon, as we said
This brings us to ask is there a reason the Moon is important to life? The answer is yes, the Earth
is tilted to its orbit by 23.5 which results in the seasons and the Moon in its orbit holds the Earth
at this angle, if it didn’t we would have wild, extreme seasons. If the tilt was less we might not
have season at all, and more the seasons might be too extreme.
K E
moon
K E
earth
(Ear th Da y) 1second
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
= 1secon d
k =
hc
2π
3
G
= 6.93E 9kg
R
H
N
A
𝔼
= 1.99E 34m
r
p
m
p
= k
R
H
N
A
𝔼
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton secon d s = carbon(C )
K E
moon
K E
earth
(Ear th Da y) 1secon d
of 32 62
Continuing On At this point we suggest the reason the Sun is in balance between its
gravitational and magnetic fields at Earth orbit for 6 proton radii is that it is to accommodate for
carbon’s central life functions. The radius of a carbon nucleus is 2.7E-15m not to be confused
with its radius, which is the distance from its center to its first electron orbital, which is
7.05E-11m or 26,000 times larger. Thus we consider equation 62
And see how many times it fits into the radius of the nucleus of carbon:
We then notice that this is the ratio of the molar mass silver (Ag) to that of gold (Au) which are
the finest electrical conductors, silver the finest at room temperature, gold the finest at extreme
temperatures, to get:
Which is exact to two places after the decimal 0.55. We take this moment to recall that the lunar
orbital radius to the solar radius is in the same as silver to gold as well, and that the Moon is
silver in color and the Sun is gold in color and that silver and gold have been used for ceremonial
jewelry since ancient times, and further recall equation 2:
We mention these things but mostly want to point out that the Moon has a function that serves
the viability of life on Earth. It is that the Earth is tilted to its orbit by 23.5 which results in the
seasons and the Moon in its orbit holds the Earth at this angle, and if it didn’t we would have
wild, extreme seasons. If the tilt was less we might not have seasons at all, and more the seasons
might be too extreme.
A picture here starts to form suggesting that life on Earth has components in the solar magnetic
field and the moon as connected with one another through electrical conductors such as the
finest gold and silver.
While gold and silver are not magnetic in any practical sense, being conductors of course with a
current run through them they can create magnetic fields, but interestingly the silicon mantle
over the core has immense amounts of gold trapped in it, enough to cover the Earth’s surface 13
inches high. It is at about 1,800 feet deep and at temperatures of many thousands of degrees.
The core is also enriched with gold and silver where temperatures are 5,400 degrees C and
pressures are up to 360 gigapascal. Actually there is enough gold in the Earth’s core to coat its
surface a foot and a half deep. The magnetic field of the earth is mostly caused by molten iron
and nickel in the Earth’s outer core; Their convective kinetic energy is converted to electric and
magnetic energy. The convection is caused by heat escaping from the core, which is a natural
process called a geodynamo.While the solar magnetic field alternates between minimum and
maximum, and flips, we may be arriving at a way to define it precisely in terms of carbon.
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
2.7E 15
4.932E 15
= 0.5474
Ag
Au
=
196.97
107.87
= 0.5476
κ
Pr otonSur fac Area
Pr oton RestEnerg y
Ear thKin et icE nerg y
Moon KineticEnerg y
=
6d a ys
LightSpeedSqu ared
of 33 62
We found the value for where the solar magnetic field balances with solar gravity results in
carbon occurring at Earth orbit, and carbon is the core element of life which is very abundant on
Earth. I have computed the values for the rest of the planets, and they don’t land on integer
values that line up with the integer values of protons in the elements. However, I did this for
Jupiter’s magnetic field in balance with its gravity for all of the Galilean satellites and they all
line-up with integer values closely. We have
Io: 14 protons. Is silicon (Si) with molar mass 28.09 grams/mole
Europa: 19 protons Is potassium (K) with molar mass 39.10 grams/mole
Ganymede: 26 protons Is iron (Fe) with molar mass 55.85 grams/mole
Callisto: 38 protons Is strontium (Sr) with molar mass 87.62 grams/mole
We see if we start with the period 2 inert gas neon (20 grams/mole) period 2 being the period
that has the most abundant life elements C, N, O and neon being an inert gas (group 18, the last
group) it determines the properties of all the elements in period 2, and the first three Galilean
satellites are produced by loop logic in computer programming. That is:
(7/5)(Neon)=(7/5)(20)=28 g/mol = Silicon = Io
We put that value in our equation (This is the loop logic):
(7/5)(Silicon)=(7/5)(28)=39.2 g/mol= Potassium = Europa
Put the result in our equation again:
(7/5)(Potassium)=(7/5)(39.2)= 54.88 = Iron =Ganymede
And, again…
(7/5)(Iron)=(7/5)(54.88)= 76.832 Is not equal to Callisto
We can write a few lines of code (Here in C) that demonstrates this:
#include <stdio.h>
int main(int argc, const char * argv[]) {
float Ne=20.00;
while (Ne<77)
{
Ne=(1.4)*Ne;
printf("%f \n", Ne);
}
return 0;
}
of 34 62
Running The Program
28.000000
39.200001
54.880001
76.832001
107.564804
Program ended with exit code: 0
All we really need is an equation for the distribution of the Galilean satellites, so we can turn the
above into an equation. And, this we can do quite nicely:
Io: 4.218E8m = 1
Europa: 6.71E8m = 1.591
Ganymede: 1.0704E9m = 2.5377
Callisto: 1.8827E9m = 4.4635
And, the equation is:
We now go on to present all of computations for all of the planets and the Galilean satellites.
a
n
= 1.6
n
a
0
= 1
a
1
= 1.6
a
2
= 2.56
a
3
= 4.096
of 35 62
Europa Orbiting Jupiter
Jupiter Magnetic Field: North 14 Gauss, South 11 Gauss
(14+11)/2=12.5 Gauss = 1.25E-3 Tesla = 12.5/0.5 = 25 Times Earth magnetic field
Jupiter mass: 1.898E27 kg
Jupiter radius: 6.9911E7 m
Europa orbits at: 6.71E8 meters
= 1.58842E-14m
1.58842E-14/0.833E-15=19.06867 proton radii = Potassium (K) almost exact 39.10 g/mol
Io orbiting Jupiter
=1.1656E-14/0.833E-15=13.993~14 proton radii = Silicon (Si) almost exact 28.09 g/mol
Ganymede Orbiting Jupiter
=2.1686E-14/0.833E-15=26.0337 proton radii~26 protons = Iron (Fe) almost exact 55.85 g/mol
Callisto Orbiting Jupiter
=3.15988E-14/0.833E-15=37.933 proton radii~38 protons = Strontium (Sr) almost exact
87.62 g/mol
19r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.898E 27kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(6.71E8m)
2
4(1.25E 3T )
2
(6.9911E 7m)
2
1/3
14r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.898E 27kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(4.218E8m)
2
4(1.25E 3T )
2
(6.9911E 7m)
2
1/3
26r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.898E 27kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.0704E 9m)
2
4(1.25E 3T )
2
(6.9911E 7m)
2
1/3
57r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.898E 27kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.8827E 9m)
2
4(1.25E 3T )
2
(6.9911E 7m)
2
1/3
of 36 62
Magnetic Field at Solar Radius
1.75E-22 m
1.75E-22/0.833E-15 = 0.000000210 proton radii
Mercury Orbit
46,00,000 to 70,000,000 kilometers
(46+70)/2=58,000,000,000 meters
=1.340E-18m
1.340E-18/0.833E-15=0.0016086 proton radii = 1 quark
Venus Orbit
=3.97347E-15m
3.97347E-15/0.833E-15=4.770 proton radii~boron = 5 protons = Boron (B)
Earth Orbit
(0.000034574)(3.47E-39)=1.19972E-43
1.19972E-43^(1/3)
=4.932E-15
4.932E-15/0.833E-15=5.92 proton radii=6 protons= Carbon (C)
0.000000210r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
4(5E 3T )
2
1/3
0.0016r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(5.8E10m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
5r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.082E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
6r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.496E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
of 37 62
Mars Orbit
=6.532E-15m
6.532E-15/0.833E-15=7.84 proton radii = 8 protons = Oxygen (O)
Asteroids
2.2-3.2 AU=2.2(1.496E11m)-3.2(1.496E11m)=3.29E11m-4.787E11m
(1.496E11+4.787E11)/2=3.1415E11 meters
=8.08794E-15m
8.08794E-15/0.833E-15=9.7094 proton radii=10 protons = Neon (Ne)
Jupiter Orbit
1.48E-14m
1.48E-14/0.833E-15=17.767 proton radii=protons= Argon (Ar)
Saturn Orbit
=2.2183E-14m/0.833E-15m=26.6308=27 protons=cobalt (Co)~Aluminum=26.98 g/mol
,
,
8r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(2.28E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
10r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(3.1415E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
18r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(7.78E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
27r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.427E12m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
= 1
gra m
atom
𝔼 =
Z gr a m s
Z protons
N
A
=
Z 6E 23protons
Z gr a m s
N
A
𝔼 = 6E 23
of 38 62
Uranus Orbit
=3.540E-14/0.833E-15=42.50 protons is between Tellurium and Iodine.
=4.77E-14/0.833E-15=57.2677 proton raddi=Lanthanum (La)
Mean Pluto Orbit
5.71454E-14/0.833E-15=68.60 proton radii=erbium (Er) which makes it outside the main block
of the periodic table in the Lanthanide series.
43r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(2.877E12m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
57r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(4.5E12m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
68r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(5.9E12m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
of 39 62
There have been five mass extinctions, and it is suspected we are experiencing the sixth right
now. We identify the times when these happened by noticing the disappearance of species in the
geologic record. The five great mass extinctions where more than 75% of the species disappeared
were:
444 million years ago, the Ordovician. 86% of species lost.
The cause was a short, severe ice age. The cause is thought to be by perhaps the uplifting of the
Appalachians, exposing silicate rock which sucked the CO2 out of the atmosphere that kept the
planet warm.
375 million years ago, The Devonian. 75% of species lost.
The cause was probably the arrival of new land plants covering the planet whose deep roots
stirred up the earth releasing nutrients into the ocean which might have caused algae blooms
that sucked oxygen out of the water suffocating the life that dwelled at the ocean bottom (for
example trilobites).
251 million years ago, The Permian. 96% of species lost
This was the worst extinction and caused by a cataclysmic eruption near Siberia which blasted
CO2 in the atmosphere, causing methanogenic bacteria to belch methane, one of the most
powerful greenhouse gases causing temperatures to rise, oceans to acidify and stagnate and
release the poisonous hydrogen sulfide.
200 million years ago, the Triassic. 200 million years ago. 80% of the species lost. They have not
been able to find a cause of this extinction.
66 million years ago, The Cretaceous. 76% of species lost.
Volcanic activity and climate change was the cause for the ammonites in the ocean as well there
was with asteroid impact that brought the dinosaur’s end.
Cosmic Rays, their story begins in 1912 when the Austrian physicist Victor Hess took a balloon
5.3 km up to measure radiation levels in the atmosphere. He found radiation levels increased
with altitude and concluded radiation was coming in from space. This radiation came to be
called “Cosmic Rays”. They are charged subatomic particles. There are two theories for their
origins: They come from massive black holes at the center of galaxies or from gamma ray bursts
(GRB’s) caused by super massive stars in distant galaxies that nova and collapse into a neutron
star or black hole. When these stars run out of fuel their radiation pressure no longer counters
gravity and the star collapses into a blackhole, and energy may be released along the axis of
rotation of the star creating a gamma ray burst. Bursts can last 10 milliseconds to several hours.
While these bursts come from distant galaxies, which must have been younger than ours when
the star went nova because the further you look, the further back in time you are looking, a
similar, weaker burst can happen in our galaxy called a soft gamma repeater which is associated
with a kind of star called a magnetar; a kind of star recently hypothesized in 1992 by Robert
Duncan and Christopher Thompson that is a type of neutron star with an extremely powerful
magnetic field (10^9 to 10^15 Gauss) which powers x-rays and gamma rays by its decay. It has
been hypothesized if one caused a gamma ray burst that pointed directly towards the Earth, it
could cause a mass extinction event. Some researchers have hypothesized such a burst was
responsible for the Ordovician mass extinction.
Interestingly, the energy source for the prebiotic path to life (prebiotic synthesis) could have
been powered by cosmic rays, (protons of 2.5 to 3.0 MeV). According to Google “High energy
of 40 62
protons are a major component of cosmic rays, and there was likely a significant flux of cosmic
rays on the early Earth.” It is becoming more thought to have been that the building blocks of
life came from ice crystals in interstellar space, or from the same in meteorites and comets. Not
just the sugar of the DNA and RNA phosphate backbone, ribose, but the nucleobases as well,
where meteorites studied have been found to have three of the five adenine, guanine, and uracil
(2022). As well amino acids that are synthesized into proteins from DNA instructions.
Recently it has been discovered in the MICMOC experiment that started in 2003 at IAS that
through a non-directed method, as opposed to methods where results are achieved by synthesis
in the laboratory, which it must be to simulate something happening of its own accord in
conditions that were thought to exist when life got its start, they found the prebiotic formation of
ribose, where prebiotic means before life and leading up to it.
They have found from studying photochemistry and thermochemistry of laboratory ices
analogous to interstellar and cometary ices, prebiotic molecules, such as amino acids in the
organic residues from sublimation of the ices. Two very important residues were discovered,
glycolaldehyde of 2 carbon atoms and glyceraldehyde of 3 carbon atoms. The sugar ribose was
produced in a formose reaction that was natural and undirected, which was a first for ribose in
that the formose reaction, discovered by Aleksandr Butlerov, was not prebiotic but rather
directed in that it required a base catalyst, where with the comet ices they used the presence of
solvated electrons that are labile in the ice acted as a catalyst. It also produces much larger
quantities of ribose. The reaction is a polymerization of formaldehyde using glycolaldehyde and
glyceraldehyde.
Back to cosmic rays. DNA and RNA are twisted ladders, a spiral helix and this makes them
chiral, but they can rotate one way or the other, left or right, and life chose one giving it a
handedness. It has been suggested that because comic rays cause mutations and thus affect
evolution and since cosmic rays produce muons, which produce electrons and positrons, that
have their magnetic moments pointing in a direction opposite their velocity, making them
polarized and always with the same handedness, and since they can interact with biological
molecules, they can give the molecules over evolutionary time the choice of one handedness over
the other, through the weak interaction. Others argue the handedness comes from local
magnetic fields, or polarized light.
Still there are those who argue, like Trevors and Abel (2004) that
“Peer-reviewed life-origin literature presupposes that, given enough time, genetic instructions
arose via natural events. Thus far, no paper has provided a plausible mechanism for natural-
process algorithm-writing. Only 200 million years separated the of Earth’s bombardment from
the presumed first appearance of life on Earth 3.8 billion years ago.”
“The genetic operating system uses a bijective coding system whereby a certain triplet codon
represents a certain amino acid…How did inanimate nature write (1) the conceptual instructions
needed to organize metabolism (2) a language/operating system needed to symbolically
represent, record and replicate those instructions? (3) a bijective coding scheme (a one-to-one
correspondence of symbol meaning) with planned redundancy so as to reduce noise pollution
between triplet codon “block code” symbols (“bytes”) and amino acid symbols? (4) How did
inanimate nature design and engineer a cell (Turing machine?) capable of implementing those
coded instructions.”
In short, Trevors and Able are raising a point that many biologists are arguing, namely, not just
that DNA can evolve through mutations, but how did a mechanism get there that can evolve, a
complete language that can encode something as complex as life, be read, and copied to
reproduce.
of 41 62
Further problems arise. Referring to Trevors and Abel describing the cell as a Turing machine,
ribosomes synthesize proteins by translating the code transcribed in mRNA into an amino acid
sequence. But they need to be synthesized themselves from amino acids, and where did the first
ribosome come from that did this?
Finally, all of that translating and describing and unzipping of DNA when it replicates, is a
process by which you would think it needs a mind to tell it what to do and control it. They have
looked for such a mind in the nucleus of cells and have not been able to find one. Further you
can say the brain does not do it because it has no nerves going from brain to the cell to the
ribosome to connect them. This is known as “The Mindless Cell Paradox.”
Stanley Miller in 1952 was working under Harold Urey, and came up with the idea of mixing
theoretical primordial elements and compounds in a bottle under electric energy to see if
biological elements would arise spontaneously. He hypothesized the early Earth atmosphere
would be composed of methane, ammonia, and hydrogen and since it had water vapor clouds
there would be lightning. The result was he produced 11 of the 20 biological amino acids. The
formose reaction is thought to parallel the reactions that lead to sugars like ribose. The Strecker
reaction is thought to parallel the reactions that produced amino acids on Earth or on meteorites
that delivered them to Earth, so-called carbonaceous chondrite meteorites.
of 42 62
of 43 62
The Cosmic Calendar Let us return to our cosmic calendar:
For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to the
moon. The days become longer by 0.0067 hours per million years. Our Equation
Is actually 1.2 seconds.
We have
24-20=0.0067t
t=597 million years
This was when the earth went through a dramatic change and there was a big explosion of life
(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went
extinct 65 million years ago giving small mammals a chance to evolve paving the way for
humans.
24-x=0.0067t
x=23.5645 hours
We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic
calendar:
What is the next term?
20+3+0.57735+0.4714=24hours
Which bring us to today. Let us see if we can make it more complete by looking at all of the five
great extinctions, not just the Cretaceous of the dinosaurs:
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
K E
m
K E
e
(Ear th Da y) = 1.2secon ds
K E
moon
K E
earth
(Ear th Da y) 1secon d
24h ours
1.2
= 20h ours
3cos(0
) +
2
3
cos(30
) = dinosaur ex t in ct ion =
20hrs + 3hrs +
3
3
hrs +
2
3
=
of 44 62
444 million years ago, the Ordovician. 86% of species lost.
The cause was a short, severe ice age. The cause is thought to be by perhaps the uplifting of the
Appalachians, exposing silicate rock which sucked the CO2 out of the atmosphere that kept the
planet warm.
375 million years ago, The Devonian. 75% of species lost.
The cause was probably the arrival of new land plants covering the planet whose deep roots
stirred up the earth releasing nutrients into the ocean which might have caused algae blooms
that sucked oxygen out of the water suffocating the life that dwelled at the ocean bottom (for
example trilobites).
251 million years ago, The Permian. 96% of species lost
This was the worst extinction and caused by a cataclysmic eruption near Siberia which blasted
CO2 in the atmosphere, causing methanogenic bacteria to belch methane, one of the most
powerful greenhouse gases causing temperatures to rise, oceans to acidify and stagnate and
release the poisonous hydrogen sulfide.
200 million years ago, the Triassic. 200 million years ago. 80% of the species lost. They have not
been able to find a cause of this extinction.
66 million years ago, The Cretaceous. 76% of species lost.
Volcanic activity and climate change was the cause for the ammonites in the ocean as well there
was an asteroid impact that brought the dinosaur’s end during this time.
Now we want to look at
t=597 million years
This was when the earth went through a dramatic change and there was a big explosion of life
(The Cambrian). This is when the Moon predicted the second as exactly 1. So we have the
starting point for the great diversification of life, the Cambrian. Our time line becomes
597 million years ago, the Cambrian
444 million years ago, the Ordovician
375 million years ago, The Devonian
251 million years ago, The Permian
200 million years ago, the Triassic
66 million years ago, The Cretaceous
Let us plot these…
of 45 62
We see the trend is 153 million years, 69 million years, 124 million years, 51 million years, 134
million years are the times from extinction event to extinction event. They alternate roughly 150,
50, 150, 50…
So, since from the Triassic the one next to the Cretaceous (dinosaurs) is 150, the next would be
50 million years after the Cretaceous plus or minus about 20 million years. The Cretaceous is at
66 million years ago so 50 plus 20 is 70 million years. We are at 66 million years since the
Cretaceous. So we are at the zone between 50 million years after the Cretaceous and 70 million
years after the Cretaceous meaning we could be, if we fit it in the graph, at extinction now plus
or minus several million years. However, we think we are in an extinction called the
Anthropocene (Man made extinction). But we can change that, we know what we need to do to
stop man-made climate change.
of 46 62
Thus we have a theoretical value for the radius of the proton:
Equation 65.
And a theoretical value for its charge
Equation 66.
Equation 67.
And we have the radius of a proton in terms of the solar magnetic field at Earth
Equation 68.
All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our
constant k
Equation 69.
And making the approximation we can with these equations eliminate in equation
4 using equation 2 in which we can eliminate (see Appendix 3 where I do this and verify the
result for both units and values and verify other equations as well that we derived earlier).
Equation 70.
This has an accuracy as shown in the appendix of close to 88% because in equation 2 the charge
of six protons is predicted by the theory to be 5.72 protons, the rest of the equations are much
more accurate, but we seek to rectify that. We write this equation so we can have an equation
that defines the solar magnetic field by solving for .
To address the accuracy of the equation for the charge of a proton, equation 2, we ask what is
the culprit. We suggest it is . So we solve the equation for that to see by how much it is
off. It is:
Equation 71.
We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%
accurate, but we want to do better (See Appendix 3).
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71pr oton s 6pr oton s
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
1 + α = 1
q
p
r
p
31850496r
2
p
=
α
4
k
e
c
4
h
π
3
G
1
(N
A
𝔼)
2
h 4π
Gc
(IE
H
)
2
μ
2
0
r
2
e
GM
m
4
p
B
2
R
2
B
(α
4
)/36
α
4
36
= q
2
k
e
c
k
2
Gc
h 4π r
2
p
of 47 62
We now eliminate on the left in equation 70 with equation 65 to find the B field of the Sun as
described by the Earth orbit as the ground state:
Equation 72.
Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives
a magnetic field strength of 4.73E-3 Teslas (See Appendix 4). Concerning equation 7
We want it to be more accurate. To do that we have to substitute for , the fraction
which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy
could be in as its experimental value has the largest errors. Other possibilities are the
is right and that as well this is due to the radius of a proton having large errors, or even that it is
supposed to the factor 8, which would be good to consider because 8-fold symmetry is very
dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear
synthesis by stars.
Because we are looking at 6 protons we had , and because of six-fold symmetry we had
and because the charge we had because of the 1/3 root on the right. So taking
and leaving the that came from substituting for k, and the as
well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation
we have:
Equation 73.
Where
Or we can factor out all the numbers on the right the 2 and 4 in equation 73 and write
Equation 74.
In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the
radius of a hydrogen atom, its ionization energy, the solar gravitational field, with the earth orbit
as the ground state.
r
2
p
B
2
=
2
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
GM
m
2
p
R
2
α
4
c
3
k
e
h 4π
Gc
α
4
36
= q
2
k
e
c
k
2
Gc
h 4π r
2
p
(α
4
)/36
7
4
5
r
p
(α
4
)/36
6
2
α
2
/6
q = 6 q
p
6
3
N = 6
7
= 279936
3
2
/4
2
8 = 2
3
B
2
=
2
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
2GM
m
2
p
4R
2
3
2
4
2
α
4
c
3
2
3
k
e
h 4π
Gc
N = 6
7
= 279936
B
2
=
1
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
GM
m
2
p
R
2
α
4
c
3
k
e
h π
Gc
r
e
of 48 62
Appendix 1
We have our equation for the Earth-Moon-Sun Orbital System (Equation 6 on page 4)
And our equation for the proton (Equations 4 and 5 on page 4)
Where is six protons is carbon (C). We also have that for the Earth orbital velocity
(Equation 16 on page 7)
This results in
Which gives
This is
Because k (Equation 18, page 7) is in s/m is 1/k=773.5 m/s. Thus we have something like a
quantum state on the left but for the planets equal to a quantum state for the proton on the
right, and conclude k connects the larger scale macrocosmos to the microcosmos of the atom. Or
perhaps better to write:
Because then we have
Because . It makes one think of how the energy of a wave is the amplitude
squared. Here amplitude then is in seconds and energy is the square of time.
K E
moon
K E
earth
(Ear th Da y) 1secon d
1
α 6m
p
h 4π r
2
p
Gc
= 1secon d
6m
p
k v
e
6
K E
moon
K E
earth
Ear th Da y =
1
α
2
k v
e
m
p
h 4π r
2
p
Gc
k
(
v
e
K E
2
moon
K E
2
earth
Ear th Da y
2
)
=
1
α
4
m
2
p
h 4π r
2
p
Gc
k (m s) = s
2
k v
e
(
K E
2
moon
K E
2
earth
Ear th Da y
2
)
=
1
α
4
m
2
p
h 4π r
2
p
Gc
s
2
= s
2
k v
e
=
s
m
m
s
= 1
of 49 62
Is actually equal to 1.2 seconds seconds squared, but we can say 1.2 seconds is a rough sketch for
the idea that it is one second, which you can do because in any physical theory there is room for
play for the physics to still serve its function. So we will evaluate it at one second. We have
On the right we have
This is an accuracy of about 95%.
Where 6 days is 6 rotations of the Earth. Just how accurate is this?
For the factor in terms of a proton we have
=
Since the mass of the earth and the mass of the moon do not vary, and the velocity of the moon
and the earth vary a little from aphelion to perihelion, their orbits are a little eccentric, we will
focus on the range of velocities.
(
K E
2
moon
K E
2
earth
Ear th Da y
2
)
k v
e
= 38.5
1
α
4
m
2
p
h 4π r
2
p
Gc
= 36.555
1
α
2
m
p
h 4π r
2
p
Gc
M
e
M
m
v
2
e
v
2
m
= 6d ays
1
α
2
m
p
h 4π r
2
p
Gc
=
18769
1.67262E 27kg
(6.62607E 34Js)(4π)(0.833E 15m)
2
(6.67408E 11Nm
2
/kg
2
)(299,792,459m /s)
(1.12213E 31)(5.37368E 31) = 6.0299675 6.03 6.0secon d s
(6.03s)
M
e
M
m
v
2
e
v
2
m
= 6d ays
M
e
= 5.972E 24kg
M
m
= 7.34767E 22kg
6d a ys = 6(24)(60)(60) = 518400secon d s
M
e
M
m
= 81.2775
(6.03s)(81.2775) = 490.103 490secon d s
of 50 62
,
The measured values are, which are mean orbital velocities of the earth and moon:
,
Thus the accuracy of the equation is 29/32.5=90%. But, using the velocity of the moon at
aphelion which is 0.966 km/s and the velocity of the earth at perihelion which is 30.29km/s, we
have 30.29/0.966=31.356. This gives us an accuracy for our equation of 31.356/32.5=96.5%.
If we take
And divide each side by , then on the left we get which is the rest energy of a proton. We
pick up then 6 days divided by the speed of light squared on the right and have equations 1 and 2
on page 3:
The rest energy of a proton is
Since kinetic energy is
We have
518400
490
= 1057.959 1058
v
2
e
v
2
m
= 1058
v
e
v
m
= 32.5
v
e
= 29.78
k m
s
v
m
= 1.022
k m
s
29.78k m /s
1.022k m /s
= 29
1
α
2
m
p
h 4π r
2
p
Gc
M
e
M
m
v
2
e
v
2
m
= 6d ays
c
2
m
p
c
2
1
α
2
h
Gc
4π r
2
p
E
p
K E
e
K E
m
6d a ys
c
2
κ
Pr otonSur fac Area
Pr oton RestEnerg y
Ear thKin et icE nerg y
Moon KineticEnerg y
=
6d a ys
LightSpeedSqu ared
κ =
1
α
2
h
Gc
= kg
s
m
= 18,769
(6.626E 34)
(6.674E 11)(299792459)
= 3.411557E 12kg
s
m
E
p
= m
p
c
2
= (1.672E 27)(299,792,459)
2
= 1.5E 10Joules
1
2
Mv
2
of 51 62
Or at perihelion
Or at aphelion
The radius of a proton is
The surface area of a proton is
We have
Where 6 days is
The accuracy of our equation is
%
Which is closer to 6 days than 5 days or 7 days and we round to the nearest integer to get 6 days.
M
e
= 5.972E 24kg
M
m
= 7.34767E 22kg
v
e
= 29,780m /s
v
e
= 30,290m /s
v
m
= 1022
v
m
= 966m /s
r
p
= 0.833E 15m
4π r
2
p
= 8.72E 30m
2
4π r
2
p
= 2.9529E 15m
5.972E 24)(30,290)
2
7.34767E 22)(966)
2
= 79,912.451
6d a ys
c
2
= (3.41E 12)
2.953E 15
1.5E 10
(79,912.451) = 5.3646E 12
6d a ys /c
2
= 6(24)(60)(60)/(8.98755E16) = = 5.768E 12
5.3646E 12
5.768E 12
100 = 93.0
of 52 62
Appendix 2
of 53 62
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Appendix 3
!
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!
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!
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!
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!
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Appendix 4
of 61 62
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The Author